Evaluate, don't forget the constant vector, ∫(t2i−2tj+1tk)dt

Question

Evaluate, don&#039;t forget the constant vector, ∫(t2i−2tj+1tk)dt

Stephanie Mann · Accepted Answer

Step 1
We have given vector-valued function as $t^{2} i - 2 t j + \frac{1}{t} k$ . While evaluating the indefinite integration of vector valued, the constant term will be considered with every component of the vector and at last, combined the constant term with their respective component.
Step 2
Now, find the integration of the vector function $t^{2} i - 2 t j + \frac{1}{t} k$ . Use the sum, difference and constant multiple rules. Also, use the power rule $\int_{}^{} u^{n} d u = \frac{u^{n + 1}}{n + 1} + C a n d \int_{}^{} \frac{1}{u} d u = \ln | u | + C$
$\int_{}^{} (t^{2} i - 2 t j + \frac{1}{t} k) d t = (\int_{}^{} t^{2} d t) i - (2 \int_{}^{} t d t) j + (\int_{}^{} \frac{1}{t} d t) k$
$= (\frac{t^{3}}{3} + C_{1}) i - (2 \cdot \frac{t^{2}}{2} + C_{2}) j + (\ln | t | + C_{3}) k$
$= \frac{t^{3}}{3} i - t^{2} j + \ln | t | k + (C_{1} i + C_{2} j + C_{3} k)$
$= \frac{t^{3}}{3} i - t^{2} j + \ln | t | k + C$
Hence, $\int_{}^{} (t^{2} i - 2 t j + \frac{1}{t} k) d t = \frac{t^{3}}{3} i - t^{2} j + \ln | t | k + C$ , where $C = C_{1} i + C_{2} j + C_{3} k$ is a constant vector.

Evaluate, don't forget the constant vector, \int_{}^{}\left(t^{2}i-

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